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Question
Prove that, if the bisector of ∠BAC of ΔABC is perpendicular to side BC, then ΔABC is an isosceles triangle.
Solution
Given: Seg AD is the bisector of ∠BAC. Seg AD ⊥ seg BC.
To prove: ΔABC is an isosceles triangle.
Proof:
In ∆ABD and ∆ACD,
∠BAD ≅ ∠CAD ...[seg AD is the bisector of ∠BAC]
seg AD ≅ seg AD ...[Common side]
∠ADB ≅ ∠ADC ...[Each angle is of measure 90°]
∴ ∆ABD ≅ ∆ACD ...[ASA test]
∴ seg AB ≅ seg AC ...[c. s. c. t.]
∴ ∆ABC is an isosceles triangle.
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